I am "puzzled" by the following puzzle from Mauricio (source: Sudopedia on "uniqueness clue cover")

- Code: Select all
`*-----------*`

|...|...|...|

|...|...|..2|

|...|..1|...|

|---+---+---|

|..1|.3.|.4.|

|..5|6.7|3..|

|.3.|.2.|..8|

|---+---+---|

|..2|.6.|5..|

|65.|..8|9.4|

|9..|4..|..7|

*-----------*

.................2.....1.....1.3..4...56.73...3..2...8..2.6.5..65...89.49..4....7 ER = 8.4

Following the very short explanation given at the sudopedia site, one can make r1c9=1, solving the puzzle with singles.

From that site:

More dramatically, if a band contains just two clues in the arrangement shown (starred) below, then all of the unclued values are eliminated from two other cells. Note that although it is obvious that -@ in r2c3 implies r2c3=1, this placement is not shown because the elimination of 2 from r2c3 does not require any uniqueness assumption: we are applying the definition strictly.

- Code: Select all
`+-------------+-------------+-------------+`

| . . . | . . . | . . *1 |

| . . -@ | . . . | . . -@ |

| . . *2 | . . . | . . . |

+-------------+-------------+-------------+

These eliminations are available because only a few special types of band (in this case) can be covered by just two clues in any proper, single-solution, puzzle. Knowing the form that these special band types take, by computer searching, allows us to make the eliminations shown.

I may have misunderstood something. Is there (nowadays) a way (proof) to justify that placement using uniqueness but without "computer searching" ? I could not find a thread on this and Wecoc's List of Acronyms points only to Sudopedia.

Many thanks in advance!